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This question arose during discussions about interesting examples of "orders of group elements" for a group theory course.

Definition: $GL(2,\mathbb{R})$ is the group of $2 \times 2$ matrices with real number entries, with non-zero determinant. The binary operation for this group is matrix multiplication.

Question: What is $\{\mathrm{ord}(M):M \in GL(2,\mathbb{R})\}$?

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    $\begingroup$ Hint: consider rotations. $\endgroup$
    – Robert Israel
    Mar 19, 2012 at 21:21
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    $\begingroup$ More interesting is the $GL(2,\mathbb{Z})$ case. $\endgroup$
    – user641
    Mar 19, 2012 at 21:22
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    $\begingroup$ Well...$GL(2, \mathbb{R})$ is a pretty big group. It contains elements of every possible order. An element of infinite order is $\left(\begin{array}{cc}1&0\\2&1\end{array}\right)$, and along with its pair $\left(\begin{array}{cc}1&2\\0&1\end{array}\right)$ they generate the free group on two generators, while one can use primitive roots of unity to find elements of any given finite order. $\endgroup$
    – user1729
    Mar 19, 2012 at 21:23
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    $\begingroup$ In order to have an element of $GL(2,{\mathbb Z})$ with order $n$ (an integer $\ge 3$), its eigenvalues would have to be primitive $n$'th roots of unity, but they are also roots of a quadratic polynomial over $\mathbb Z$. Since the $n$'th cyclotomic polynomial is irreducible over the rationals, the only possibilities are $n=3,4,6$ which are the cases where the cyclotomic polynomials are quadratic. $\endgroup$
    – Robert Israel
    Mar 19, 2012 at 21:36
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    $\begingroup$ So you can have $\pmatrix{1&0\cr 0 & 1\cr}$ of order $1$, $\pmatrix{-1&0\cr 0 & 1\cr}$ of order $2$, $\pmatrix{0 & -1\cr 1 & -1\cr}$ of order $3$, $\pmatrix{0 & -1\cr 1 & 0\cr}$ of order $4$, or $\pmatrix{0 & -1\cr 1 & 1\cr}$ of order $6$. $\endgroup$
    – Robert Israel
    Mar 19, 2012 at 21:40

1 Answer 1

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It can be any number. Take a matrix that represents a rotation with an angle $\phi=\frac{2\pi}{n}$ ($n\in \mathbb{Z}$), that is $$\left(\begin{array}{cc}\cos\phi & \sin \phi\\-\sin\phi & \cos\phi\end{array}\right)$$ Its order is clearly $n$.

EDIT: Just for completeness, it can of course also be infinite. An example would be $$\left(\begin{array}{cc}1 & 1\\ 0 & 1\end{array}\right)$$ which is invertible and satisfies $$\left(\begin{array}{cc}1 & 1\\ 0 & 1\end{array}\right)^n=\left(\begin{array}{cc}1 & n\\ 0 & 1\end{array}\right)$$

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  • $\begingroup$ Thanks for that. That very succinctly answers my question. $\endgroup$
    – Douglas S. Stones
    Mar 19, 2012 at 21:29

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